A) hexanamine
B) propanamme
C) butanamine
D) pentanamme
Correct Answer: D
Solution :
When amide is heated with a mixture of\[B{{r}_{2}}\]in the presence of\[NaOH\]or KOH, amine is formed which has one carbon atom less than original amide. This is called Hofmanns degradation reaction. \[Hexanamide+B{{r}_{2}}+4KOH\xrightarrow{{}}Pentanamine\] \[+{{K}_{2}}C{{O}_{3}}+2KBr+2{{H}_{2}}O\]
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